Mathematics: Post your doubts here!

[Igcse stuff]

Thanks a lot for the detailed explantions!!! ishaallah

 

[David Hussey]

thankyou! any clue of the question?

can u give me the link to the question?

 

[salvatore]

Hey,, no worries. Glad 2 help.

So, 9(ii):
I guess you know what we gotta do in this - we treat the gradient of normal as a line. To find the stationary (max) point on it, we find the dy/dx of this 'line'. (Technically we're findin the d^2y/dx^2)

From (i), we know dy/dx is (1+x).[(1-x^2)]^(1/2)___________ [Here, (1+x) is u and v is [(1-x^2)]^(1/2)]
Apply (u.v)' = u'.v+u.v'
d^2y/dx^2 = (1).[(1-x^2)]^(1/2) + (1+x).(1/2)[(1-x^2)]^(-1/2)(-2x)
= [(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2)
Now equate this to zero:
[(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2) = 0
I then got: (1-x^2) - x(1+x) = 0 ___________ (Coz [(1-x^2)]^(1/2).[(1-x^2)]^(1/2) = (1-x^2) )
Then: 1 - x^2 - x - x^2 = 0
-2x^2 - x + 1 = 0 (multiply with -1 in both sides)
2x^2 + x - 1 = 0
Solving this you get two sol., x = 1/2 (ANS.) or x = -1
From graph, we see dat point P ain't on x = -1
Ans. is thus x = 1/2

damn.. I had made a small error in differentiating.
Thanks a lot man!

 

[Obsidian Fl1ght]

damn.. I had made a small error in differentiating.
Thanks a lot man!

No prob!

 

[Ruchi1307]

solve the question with steps please..

 

[Suchal Riaz]

my goodness...

Sorry but i don't seem to understand the reason behind your amazement?

 

[Suchal Riaz]

solve the question with steps please..

 

[David Hussey]

Sorry but i don't seem to understand the reason behind your amazement?

no dude its soo complicated!
btw m doing p3 as well... and relli scared about p3!

 

[David Hussey]

View attachment 36464

u must be a math genius! mashAllah!

 

[Suchal Riaz]

no dude its soo complicated!
btw m doing p3 as well... and relli scared about p3!

Well u might be amazed to hear that i am in as level and i did P3 for curiosity just once a few months back.

 

[David Hussey]

Well u might be amazed to hear that i am in as level and i did P3 for curiosity just once a few months back.

:O

 

[Suchal Riaz]

:O

If u have any other question send me here on this thread and tag me. Anything except s1.

 

[ahmed abdulla]

If u have any other question send me here on this thread and tag me. Anything except s1.

q-9(ii)
complex no.

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_32.pdf
Please help me with qn no. 9(ii).
Thanks

 

[Suchal Riaz]

q-9(ii)
complex no.

I hate this topic really.

 

[ahmed abdulla]

View attachment 36478I hate this topic really.

thats also my worst toppic .
Thanks bro

 

[Suchal Riaz]

thats also my worst toppic .
Thanks bro

Actually it is not difficult but tedious. No problem. Feel free to post any question. It will be like practice to me as i have'nt done past papers

 

[ahmed abdulla]

question
7 (ii)

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_32.pdf
Please help me with qn no. 9(ii).
Thanks

 

[Suchal Riaz]

question
7 (ii)

I am just guessing btw. For y to be zero, either sin or cos have to be zero. Graph starts at zero as sin is zero. Then graph comes to zero when cos is zero. Which is at 2X=pi/2. Therefore x=pi/4. For one area the interval is of pi/4. For 40 A it must be 40/4=10pi. Is my answer correct?